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In page 64 of Bayesian Data Analysis by Gelman et.al. they write

... sensible vague prior density for µ and σ, assuming prior independence of location and scale parameters, is uniform on ($\mu$, $\log~\sigma$) or, equivalently, $p(\mu, \sigma^2) \propto 1/\sigma^2$.

Also in page 87 of The BUGS Book (pdf download) they discuss the equivalence of the Jeffreys prior to the uniform on the scale:

... the Jeffreys prior is $p_J(\sigma) \propto \sigma^{-1}$, which in turn means that $p_J(\sigma^k) \propto \sigma^{-k}$ for any choice of power k. ... we note that the Jeffreys prior is equivalent to $p_J(log \sigma^k) \propto constant$.

I have understood this to mean that a uniform prior on $\log \sigma^2$ should be $\propto 1/\sigma^2$. I have been unable to derive this. This is my attempt (with a nod to this answer):

\begin{align} \text{Let}~ Y =& \log \sigma^2 \\ p(Y) \propto& 1 \\ \frac{dY}{d\sigma^2} =& 2/\sigma \end{align}

Then to get the distribution on the $\sigma^2$ scale:

\begin{align} \text{If}~ X =& \sigma^2 \\ \text{then}~ p(X) =& p(Y) |\frac{dY}{d\sigma^2}| \\ =& 1 \times 2/\sigma \\ \propto & 1/\sigma \end{align}

Where have I gone wrong please?

user2957945
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1 Answers1

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Your error is going from $\text{Let}~ Y = \log \sigma^2$ to $\dfrac{dY}{d\sigma^2} = 2/\sigma$

You should have: $\dfrac{dY}{d\sigma^2} = 1/{\sigma^2}$ (simple derivative of a logarithm)

though perhaps you tried $\dfrac{dY}{d\sigma} = 2\sigma \frac{1}{\sigma^2}= 2/{\sigma}$ (chain rule).

This gives you: $\text{If}~ X = \sigma^2 \text{ then } p(X) \propto 1/{\sigma^2}$ for the improper prior.

It is worth noting that $\log(\sigma^k)=k\log(\sigma)$ and this indicates why something similar happens for all powers of the standard deviation.

Henry
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